QJEEM23Mae365: JEE Main 2023 Mathematics - Class 11 JEEM23Mae365 Solution: Let and be four points on the ellipse . Let and be mutually…

Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\frac{x^2}{4} + \frac{y^2}{9} = 1$ . This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$ -axis and semi-minor axis $b=2$ along the $x$ -axis.

OP is the distance from origin O to point P, which is given by :

$OP =r_1 = \sqrt{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{12}{7} + \frac{36}{7}} = \sqrt{\frac{48}{7}} = 2\sqrt{\frac{12}{7}}.$

1. Let's represent the given point $P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$ in polar coordinates. We can write $P$ as $(r_1 \cos \theta, r_1 \sin \theta)$ . Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1 \cos \theta$ and $y=r_1 \sin \theta$ into the equation of the ellipse gives us :

$\frac{r_1^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1$

Simplifying this, we obtain :

$\frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48} \quad \text{--- (equation 1)}$

2. Similarly, if we represent the point $R$ as $(-r_2 \sin \theta, r_2 \cos \theta)$ , (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\pi/2$ radians. In terms of sin and cos, $\sin(\theta + \pi/2) = \cos(\theta)$ and $\cos(\theta + \pi/2) = -\sin(\theta)$ .) it too should satisfy the equation of the ellipse. We have :

$\frac{r_2^2 \sin ^2 \theta}{4}+\frac{r_2^2 \cos ^2 \theta}{9}=1$

Simplifying this, we obtain :

$\frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \quad \text{--- (equation 2)}$

3. From equations (1) and (2), we have :

$\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$

4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ = 2OP$ and $RS = 2OR$ . Thus,

$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} \right)$

5. Substituting the values of $r_1$ and $r_2$ , we obtain :

$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144} = \frac{p}{q}$

6. Hence, $p = 13$ and $q = 144$ .

7. So, the final answer $p+q = 13 + 144 = 157$ .